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Q4(vi):
Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (vi) $\sqrt{\frac{1 + \sin A}{1 – \sin A}} = \sec A + \tan A$
Solution :
Given: An identity involving an acute angle $A$, specifically $\sqrt{\frac{1 + \sin A}{1 – \sin A}}$.
To Prove: $\sqrt{\frac{1 + \sin A}{1 – \sin A}} = \sec A + \tan A$
Step 1: Consider the Left Hand Side (LHS) of the identity.
LHS = $\sqrt{\frac{1 + \sin A}{1 - \sin A}}$
Step 2: Rationalize the denominator inside the square root.
To rationalize the denominator, we multiply the numerator and the denominator by the conjugate of the denominator, which is $(1 + \sin A)$.
LHS = $\sqrt{\frac{(1 + \sin A) \times (1 + \sin A)}{(1 - \sin A) \times (1 + \sin A)}}$
Step 3: Simplify the expression using algebraic identities.
In the numerator, we have $(1 + \sin A)(1 + \sin A) = (1 + \sin A)^2$.
In the denominator, we use the difference of squares identity: $(a - b)(a + b) = a^2 - b^2$.
Here, $(1 - \sin A)(1 + \sin A) = 1^2 - \sin^2 A = 1 - \sin^2 A$.
LHS = $\sqrt{\frac{(1 + \sin A)^2}{1 - \sin^2 A}}$
Step 4: Apply the fundamental trigonometric identity.
We know that $\sin^2 A + \cos^2 A = 1$, which implies that $1 - \sin^2 A = \cos^2 A$.
Substituting this into our expression:
LHS = $\sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}}$
Step 5: Extract the square root.
Since the square root of a squared term is the term itself (given $A$ is an acute angle, $\cos A > 0$ and $1 + \sin A > 0$):
LHS = $\frac{1 + \sin A}{\cos A}$
Step 6: Separate the terms in the fraction.
LHS = $\frac{1}{\cos A} + \frac{\sin A}{\cos A}$
Step 7: Apply trigonometric reciprocal and quotient identities.
We know that $\frac{1}{\cos A} = \sec A$ and $\frac{\sin A}{\cos A} = \tan A$.
LHS = $\sec A + \tan A$
Conclusion:
Since the Left Hand Side (LHS) is equal to the Right Hand Side (RHS), the identity is proven.
Final Answer: $\sqrt{\frac{1 + \sin A}{1 – \sin A}} = \sec A + \tan A$ is proved.
More Questions from Class 10 Mathematics Introduction to Trigonometry EXERCISE 8.3
- Q1: Express the trigonometric ratios $\sin A$, $\sec A$ and $\tan A$ in terms of $\cot A$.
- Q2: Write all the other trigonometric ratios of $\angle A$ in terms of $\sec A$.
- Q3(i): Choose the correct option. Justify your choice. (i) $9 \sec^2 A – 9 \tan^2 A =$
- Q3(ii): Choose the correct option. Justify your choice. (ii) $(1 + \tan \theta + \sec \theta) (1 + \cot \theta – \text{cosec } \theta) =$
- Q3(iii): Choose the correct option. Justify your choice. (iii) $(\sec A + \tan A) (1 – \sin A) =$
- Q3(iv): Choose the correct option. Justify your choice. (iv) $\frac{1 + \tan^2 A}{1 + \cot^2 A} =$
- Q4(i): Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (i) $(\text{cosec } \theta – \cot \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}$
- Q4(ii): Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (ii) $\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2 \sec A$
- Q4(iii): Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (iii) $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \text{cosec } \theta$ [Hint : Write the expression in terms of $\sin \theta$ and $\cos \theta$]
- Q4(iv): Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (iv) $\frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1 – \cos A}$ [Hint : Simplify LHS and RHS separately]
- Q4(ix): Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (ix) $(\text{cosec } A – \sin A) (\sec A – \cos A) = \frac{1}{\tan A + \cot A}$ [Hint : Simplify LHS and RHS separately]
- Q4(v): Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (v) $\frac{\cos A – \sin A + 1}{\cos A + \sin A – 1} = \text{cosec } A + \cot A$, using the identity $\text{cosec}^2 A = 1 + \cot^2 A$.
- Q4(vii): Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (vii) $\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta$
- Q4(viii): Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (viii) $(\sin A + \text{cosec } A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$
- Q4(x): Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (x) $(\frac{1 + \tan^2 A}{1 + \cot^2 A}) = (\frac{1 - \tan A}{1 - \cot A})^2 = \tan^2 A$
CBSE Solutions for Class 10 Mathematics Introduction to Trigonometry
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